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The area of a pyramid with a square base with side length $L$ and height $h$ is calculated as follows:

  1. In $\mathbb{R}^3$ place the pyramid upright with one side flush with the z-axis so that the corners of the base are at $(0,0,\frac{1}{L})$, $(0,0,-\frac{1}{L})$, $(L,0,\frac{1}{L})$ and $(L,0,-\frac{1}{L})$. alt text

  2. We will slice parallel to the yz-plane from $x=0$ up to $x=\frac{L}{2}$. (We can then double it to get the whole pyramid.) The slices are rectangles that all have a base equal to L and height given by $f(x)=\frac{2hx}{L}$. The cross-sectional area is then, $A(x) = L\frac{2hx}{L}=2hx$.

  3. So, we have $2 \int_{0}^{\frac L2} (2hx) dx \neq \frac13 L^2 h$.

What has gone wrong?

ANS: The slices are NOT rectangles. DUH. Thanks, Rahul Narain.

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I don't see how you are placing your pyramid. "One side flush with the $z$-axis" is not enough information, as one may rotate the shape about the $z$-axis while keeping it flush. If the base lies on the $xz$-plane, its corners have $x$-coordinate $0$ or $L$, not $1/L$, and the slices parallel to the $yz$-plane will not be rectangles but trapeziums (a.k.a. trapeziods), as one can see by considering the slice at $L/2$ which is clearly a triangle (a degenerate trapezium) and not a rectangle. –  Rahul Nov 16 '10 at 6:48
    
"If the base lies on the xz-plane, its corners have x-coordinate 0 or L." That's what I have. My co-ordinates are (x,y,z). But you are right about the trapezoid. I have no idea why I didn't see that. –  a little don Nov 16 '10 at 6:51
    
@a little don: Whoops, I meant the $z$-coordinates. Sorry, it's late in my time zone. –  Rahul Nov 16 '10 at 6:56
    
It's late here too. I'm still mad I didn't see the trapezoids. Thanks. I feel sane again. It works when I use the trapezoid area formula. –  a little don Nov 16 '10 at 6:59
    
Considering your meta thread @Rahul, would you mind posting that as an answer? ;) –  J. M. Nov 16 '10 at 7:07

1 Answer 1

up vote 2 down vote accepted

This is an elaboration of my comment, so the question can be marked as answered.

I don't see how you are placing your pyramid. "One side flush with the z-axis" is not enough information, as one may rotate the shape about the $z$-axis while keeping it flush. If the base lies on the $xz$-plane, its corners have $z$-coordinate $0$ or $L$, not $1/L$, and the slices parallel to the $yz$-plane will not be rectangles but trapeziums (a.k.a. trapeziods), as one can see by considering the slice at $L/2$ which is clearly a triangle (a degenerate trapezium) and not a rectangle.

A slice at a given $x$-coordinate $x < L/2$ is a trapezium with height $2hx/L$, as you noticed, but with one parallel size of length $L$ and the other of length $L-2x$. The area of the trapezium is therefore $A = 2hx/L\cdot(L - x) = 2hx\cdot(1-x/L)$, and the volume of the pyramid is $V = 2\int_{x=0}^{L/2} A dx = hL^2/3$ as expected.

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